5 Marks Questions

Question 15 Marks

The following table gives the life time of 400 neon lamps:

Life time(in hours)	300 - 400	400 - 500	500 - 600	600 - 700	700 - 800	800 - 900	900 -1000
Number of lamps:	14	56	60	86	74	62	48

A bulb is selected at random. Find the probability that the lifetime of a selected bulb is:

Less than 400
between 300 to 800 hours
At least 700 hours

Answer

Total number of bulbs = 400

Probability that the life of the selected bulb is less than 400 hrs

$=\frac{\text{No. of bulbs having life less than 400 hrs}}{\text{Total no. of bulbs}}$
$=\frac{14}{400}$
$=\frac{7}{200}$

Probability that the life of the selected bulb is between 300 - 800 hrs

$=\frac{\text{No. of bulbs having life less than 400 hrs}}{\text{Total no. of bulbs}}$
$=\frac{14+56+60+86+74}{200}$
$=\frac{290}{400}$
$=\frac{29}{40}$

Probability that the life of the selected bulb is at least 700 hrs

$=\frac{\text{No. of bulbs having life less than 700 hrs}}{\text{Total no. of bulbs}}$
$=\frac{74+62+48}{400}$
$=\frac{184}{400}$
$=\frac{23}{50}$

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Question 25 Marks

Define probability of an event.

Answer

The probability of an event denotes the relative frequency of occurrence of an experiment’s outcome, when repeating the experiment. Definition: The empirical or experimental definition of probability is that if n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials, then the probability of happening of event A is denoted by P(A) and is given by $\text{P(A)}=\frac{\text{m}}{\text{n}}$ To illustrate the definition, let us take examples:

When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT. The total number of trials is 4. Let A be the event of occurring exactly two heads. The number of times A happens is 1. So, the probability of the event A is

$\text{P(A)}=\frac{\text{m}}{\text{n}}$
$=\frac{1}{4}$
$=0.25$

In the experiment of rolling a dice, the possible outcomes are 1, 2, 3, 4, 5 and 6. Let A be the event of occurring a number greater than 3. The total number of trials is 6. The number of times A happens is 3. So, the probability of the event A is

$\text{P(A)}=\frac{\text{m}}{\text{n}}$
$=\frac{3}{6}$
$=\frac{1}{2}$
$=0.5$
Note that H stands for getting a head and T stands for getting a tail in the experiment of tossing a coin.

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Question 35 Marks

1500 families with 2 children were selected randomly, and the following data were recorded:

No of girls in a family	0	1	2
No of girls	211	814	475

If a family is chosen at random, compute the probability that it has:

No girl.
1 girl.
2 girls.
At most one girl.
More girls than boys.

Answer

Probability of having no girl in a family $=\frac{\text{No. of families having no girl}}{\text{Total no. of families}}$

$=\frac{211}{1500}=0.1406$

Probability of having 1 girl in a family $=\frac{\text{No. of families having 1 girl}}{\text{Total no. of families}}$

$=\frac{814}{1500}$
$=\frac{407}{750}=0.5426$

Probability of having 2 girl in a family $=\frac{\text{No. of families having 2 girl}}{\text{Total no. of families}}$

$=\frac{475}{1500}=0.3166$

Probability of having at the most one girl $=\frac{\text{No. of families having at most one girl}}{\text{Total no. of families}}$

$=\frac{211+814}{1500}$
$=\frac{1025}{1500}=0.6833$

Probability of having more girls than boys $=\frac{\text{No. of families having more girls than boys}}{\text{Total no. of families}}$

$=\frac{475}{1500}=0.31$

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Question 45 Marks

Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:

Outcome	No head	One head	Two head	Three head
Frequency	14	38	36	12

If the three coins are tossed simultaneously again, compute the probability of:

2 heads coming up.
3 heads coming up.
At least one Head coming up.
Getting more Tails than Heads.
Getting more heads than tails.

Answer

Probability of 2 heads coming up $=\frac{\text{Favorable out come}}{\text{Total out come}}$

$=\frac{36}{100}=0.36$

Probability of 3 heads coming up $=\frac{\text{Favorable out come}}{\text{Total out come}}$

$=\frac{12}{100}=0.12$

Probability of at least one head coming up $=\frac{\text{Favorable out come}}{\text{Total out come}}$

$=\frac{38+36+12}{100}$
$=\frac{86}{100}=0.86$

Probability of getting more heads than Tails $=\frac{\text{Favorable out come}}{\text{Total out come}}$

$=\frac{36+12}{100}$
$=\frac{48}{100}=0.48$

Probability of getting more tails than heads $=\frac{\text{Favorable out come}}{\text{Total out come}}$

$=\frac{14+38}{100}$
$=\frac{52}{100}=0.52$

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